arctanx的原函数
发布网友
共2个回答
热心网友
令a=1，原式=
x arctanx - (1/2)ln(1+x²) + C
热心网友
∫ arctanx dx
= x * arctanx - ∫ x d(arctanx)
= x * arctanx - ∫ x/(1+x²) dx
= x * arctanx - (1/2)∫ d(x²)/(1+x²)
= x * arctanx - (1/2)∫ d(1+x²)/(1+x²)
= x * arctanx - (1/2)ln(1+x²) + C
